Potential superposition and electric dipoles

Superposition for potential is inherited from superposition for force. Put a test charge q₀ at P. Each source charge pushes on it independently, so the work the electrostatic force does as q₀ travels from P out to infinity is the sum of the works done by each source separately. Divide that total by q₀ and you have V(P) = k Σ qᵢ/rᵢ, where rᵢ is the distance from charge i to P and each qᵢ enters with its own sign. No component is taken, no angle appears, and the order of the terms is irrelevant. Compare the cost: five charges give five signed numbers to add for V, against five vectors and fifteen components for E. That is why potential is the first quantity to compute whenever more than two charges are present, and why every term must be written against the same reference — here V = 0 at infinity, the convention that makes k q/r the potential of a single charge.

Take q₁ = +8.0 nC at the origin and q₂ = −2.0 nC at x = 0.60 m, and ask where V = 0 on the x-axis. Setting k q₁/r₁ + k q₂/r₂ = 0 cancels k and leaves a pure ratio condition, 8.0/r₁ = 2.0/r₂, so r₁ = 4r₂: the point must be four times as far from the large charge as from the small one. Between the charges r₁ = x and r₂ = 0.60 − x, so x = 4(0.60 − x) gives x = 0.48 m. Beyond the small charge r₂ = x − 0.60, so x = 4(x − 0.60) gives x = 0.80 m. Two zeros, both nearer the weaker charge, and none on the far side of the stronger one, where the +8.0 nC term always wins. Notice what the condition never involved: no directions and no angles, only distances and signs, so each region needed one linear equation.

Put +q at z = +d/2 and −q at z = −d/2, and look from a point a distance r from the centre at an angle θ to the +z axis. Exactly, V = k q (1/r₊ − 1/r₋) = k q (r₋ − r₊)/(r₊ r₋). Both terms are individually large and they almost cancel, so the answer lives entirely inside the small difference r₋ − r₊. The cosine rule gives r∓ = √(r² ∓ r d cos θ + d²/4), and for r ≫ d the binomial expansion keeps only the first correction, r∓ ≈ r(1 ∓ (d/2r) cos θ). Then r₋ − r₊ ≈ d cos θ while the product r₊ r₋ ≈ r², so V = k q d cos θ/r², or V = k p cos θ/r² with p = qd. The derivation carries its own warning: you may not expand each term to leading order on its own, because 1/r₊ ≈ 1/r and 1/r₋ ≈ 1/r would predict V = 0 everywhere.

Two features come out of that formula and neither belongs to a single charge. First, the falloff is one power faster: 1/r² rather than 1/r. A water molecule has p = 6.2 × 10⁻³⁰ C m, so on its axis V = k p/r² is 56 mV at r = 1.0 nm and 0.56 mV at 10 nm, while a lone proton's potential only falls from 1.44 V to 0.144 V across the same stretch — the dipole gives away another factor of ten every decade. That is what neutrality costs: the monopole term k(Σq)/r is exactly zero, and the dipole term is the first survivor. Second, the potential carries an angle. It is positive in the hemisphere p points into, negative behind, largest in magnitude on the axis where cos θ = ±1, and zero at θ = 90°. Walk around a sphere of fixed r and V changes sign, which no single charge can do.

Solving V = 0 along a line gave points; in three dimensions it gives surfaces. For a dipole the condition is cos θ = 0, so V vanishes on the entire plane that perpendicularly bisects the pair — and that result is exact rather than far-field, because every point of the plane has r₊ = r₋ and the two terms cancel identically. When the magnitudes differ the surface closes up. For the +8.0 nC and −2.0 nC pair the condition r₁ = 4r₂ is met on a sphere, the locus of points whose distances to two fixed points hold a constant ratio. Its axis crossings are the 0.48 m and 0.80 m already found, so it is centred 0.64 m from the large charge with radius 0.16 m — wrapped round the weaker charge, never round the stronger. A zero-potential surface is an equipotential like any other, and where it sits depends on putting V = 0 at infinity.

V = 0 is a statement about a number, not about a slope. On the dipole's bisecting plane V is zero everywhere in the plane, so V has no rate of change within it and E has no in-plane component; but step off the plane either way and V changes immediately, so E is perpendicular to the plane, of magnitude k p/r³ and pointing antiparallel to p. A charge released there does not stay. The two conditions separate in the other direction too. For the +8.0 nC and −2.0 nC pair the field vanishes at x = 1.20 m, where 8.0/x² = 2.0/(x − 0.60)², and the potential there is (8.99 × 10⁹)(8.0 × 10⁻⁹/1.20 − 2.0 × 10⁻⁹/0.60) = +30 V. Meanwhile at x = 0.48 m, where V = 0, the two field contributions both point along +x and sum to 1.56 × 10³ N C⁻¹. Neither condition implies the other.